Question

If the shortest distance between the lines \(\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{-3} and \frac{x - \lambda}{2} = \frac{y + 1}{4} = \frac{z - 2}{-5}\) is \(\frac{6}{\sqrt{5}}\), then the sum of all possible values of \(\lambda\) is:

• A. \( 5 \)
• B. \( 8 \)
• C. \( 7 \)
• D. \( 10 \)

Answer 1

The Correct Option is B

To find the sum of all possible values of \(\lambda\) for which the shortest distance between the given lines is \(\frac{6}{\sqrt{5}}\), we proceed as follows:

The given lines are: 

• First line: \(\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{-3}\)
• Second line: \(\frac{x - \lambda}{2} = \frac{y + 1}{4} = \frac{z - 2}{-5}\)

The general form of the equations of the lines are:

• First line in vector form: \(\mathbf{r}_1 = \langle 4, -1, 0 \rangle + t \langle 1, 2, -3 \rangle\)
• Second line in vector form: \(\mathbf{r}_2 = \langle \lambda, -1, 2 \rangle + s \langle 2, 4, -5 \rangle\)

The shortest distance \(d\)\) between two skew lines \(\mathbf{r}_1 = \mathbf{a}_1 + t\mathbf{b}_1\) and \(\mathbf{r}_2 = \mathbf{a}_2 + s\mathbf{b}_2\) is given by:

• \(d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}\)

Here, \(\mathbf{a}_1 = \langle 4, -1, 0 \rangle\), \(\mathbf{b}_1 = \langle 1, 2, -3 \rangle\), \(\mathbf{a}_2 = \langle \lambda, -1, 2 \rangle\), and \(\mathbf{b}_2 = \langle 2, 4, -5 \rangle\).

First compute the cross product \((\mathbf{b}_1 \times \mathbf{b}_2)\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \mathbf{i}(2 \times -5 - (-3)\times 4) - \mathbf{j}(1\times -5 -(-3)\times 2) + \mathbf{k}(1\times 4 - 2\times 2)\)

\(= \mathbf{i}(-10 + 12) - \mathbf{j}(-5 + 6) + \mathbf{k}(4 - 4)\)

\(= \mathbf{i}(2) - \mathbf{j}(1) + \mathbf{k}(0)\)

\(= \langle 2, -1, 0 \rangle\)

Now, compute \((\mathbf{a}_2 - \mathbf{a}_1)\):

\(= \langle \lambda - 4, 0, 2 \rangle\)

Compute the dot product \((\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)\):

\(= (\langle \lambda - 4, 0, 2 \rangle) \cdot \langle 2, -1, 0 \rangle\)

\(= 2(\lambda - 4) + 0\right)\)

\(= 2\lambda - 8\)

The magnitude of \((\mathbf{b}_1 \times \mathbf{b}_2)\): \(|\langle 2, -1, 0 \rangle| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}\)

Setting up the equation for the shortest distance:

\(\frac{|2\lambda - 8|}{\sqrt{5}} = \frac{6}{\sqrt{5}}\)

Therefore, \(|2\lambda - 8| = 6\)

This gives two equations:

• \(lambda - 8 = 6\) \(\Rightarrow 2\lambda = 14\) \(\Rightarrow \lambda =\)
• \(lambda - 8 = -6\) \(\Rightarrow 2\lambda = 2\) \(\Rightarrow \lambda =\)

The possible values of \(\lambda\) are \(7\) and \(1\). Therefore, the sum of all possible values is:

Sum = \(7 + 1 = 8\)

Thus, the sum of all possible values of \(\lambda\) is \(8\).

Answer 2

Given:
\(\int_0^1 \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} \, dx = a + b \sqrt{2} + c \sqrt{3}\)

where \(a, b, c\) are rational numbers.

Step 1. Simplifying the Integral: Consider
\(\int_0^1 \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} \, dx\)

Rationalizing the denominator:
\(\int_0^1 \frac{\sqrt{3 + x} - \sqrt{1 + x}}{(3 + x) - (1 + x)} \, dx = \int_0^1 \frac{\sqrt{3 + x} - \sqrt{1 + x}}{2} \, dx\)

  Therefore:
 \(= \frac{1}{2} \int_0^1 \left( \sqrt{3 + x} - \sqrt{1 + x} \right) \, dx\)

Step 2. Separating the Integral:
 \(= \frac{1}{2} \left( \int_0^1 \sqrt{3 + x} \, dx - \int_0^1 \sqrt{1 + x} \, dx \right)\)
Step 3. Evaluating Each Integral:
- For   \(\int_0^1 \sqrt{3 + x} \, dx \:\)
  \(\int \sqrt{3 + x} \, dx = \frac{2}{3} (3 + x)^{3/2}\)
     Evaluating from 0 to 1:

    \(\frac{2}{3} \left( (3 + x)^{3/2} \right) \Big|_0^1 = \frac{2}{3} \left( (4)^{3/2} - (3)^{3/2} \right) = \frac{2}{3} (8 - 3\sqrt{3})\)
  

  For\(\int_0^1 \sqrt{1 + x} \, dx\):
   
    \(\int \sqrt{1 + x} \, dx = \frac{2}{3} (1 + x)^{3/2}\)
 

    Evaluating from 0 to 1:
\(\frac{2}{3} \left( (1 + x)^{3/2} \right) \Big|_0^1 = \frac{2}{3} \left( (2)^{3/2} - (1)^{3/2} \right) = \frac{2}{3} (2\sqrt{2} - 1)\)

Step 4. Combining the Results:

  \(\frac{1}{2} \left( \frac{2}{3} (8 - 3\sqrt{3}) - \frac{2}{3} (2\sqrt{2} - 1) \right)\)

  Simplifying:

 \(\frac{1}{3} (8 - 3\sqrt{3} - 2\sqrt{2} + 1) = \frac{1}{3} (9 - 3\sqrt{3} - 2\sqrt{2})\)

Thus:

\(a = 3, \quad b = -\frac{2}{3}, \quad c = -1\)
 

Step 5. Calculating \( 2a + 3b - 4c \):
\(2a + 3b - 4c = 2 \times 3 + 3 \times \left( -\frac{2}{3} \right) - 4 \times (-1)\)
\(= 6 − 2 + 4 = 8\)